Q90 / 125 · BITSAT 2014 · Physics
A person with his hand in his pocket is skating on ice at the rate of $10 \mathrm{~m} / \mathrm{s}$ and describes a circle of radius $50 \mathrm{~m}$. What is his inclination to vertical:
$\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{sec}^{2}\right)$
- A$\tan ^{-1}(1 / 2)$
- Correct answerB$\tan ^{-1}(1 / 5)$
- C$\tan ^{-1}(3 / 5)$
- D$\tan ^{-1}(1 / 10)$
Solution
Since surface (ice) is frictionless, so the centripetal force required for skating will be provided by inclination of boy with the vertical and that angle is given as $\tan \theta=\frac{\mathrm{v}^{2}}{\mathrm{rg}}$ where $\mathrm{v}$ is speed of skating $\&$
i is radius of circle in which he moves.
Hence, the correct answer is option: 'B'.